1. **State the problem:** We have three pipes filling a tank with different rates and a leak draining the tank. We want to find the total time to fill the tank completely. 2. **Given data:** - Pipe A fills the tank in 5 hours, so its rate is $\frac{1}{5}$ tank/hour. - Pipe B fills the tank in 6 hours, so its rate is $\frac{1}{6}$ tank/hour. - Pipe C fills the tank in 8 hours, so its rate is $\frac{1}{8}$ tank/hour. - Leak drains $\frac{1}{10}$ tank/hour. - Pipe B shuts off after 2 hours. - Pipe C shuts off 30 minutes (0.5 hours) after Pipe B, i.e., after 2.5 hours. 3. **Approach:** We calculate the amount of water filled in each time interval considering the pipes active and the leak. 4. **Interval 1: From 0 to 2 hours (all pipes on)** - Combined filling rate: $\frac{1}{5} + \frac{1}{6} + \frac{1}{8} = \frac{24}{120} + \frac{20}{120} + \frac{15}{120} = \frac{59}{120}$ tank/hour. - Leak rate: $\frac{1}{10} = \frac{12}{120}$ tank/hour. - Net filling rate: $\frac{59}{120} - \frac{12}{120} = \frac{47}{120}$ tank/hour. - Water filled in 2 hours: $$2 \times \frac{47}{120} = \frac{94}{120} = \frac{47}{60}$$ tank. 5. **Interval 2: From 2 to 2.5 hours (Pipe B off, Pipes A and C on)** - Filling rate: $\frac{1}{5} + \frac{1}{8} = \frac{8}{40} + \frac{5}{40} = \frac{13}{40}$ tank/hour. - Leak rate: $\frac{1}{10} = \frac{4}{40}$ tank/hour. - Net filling rate: $\frac{13}{40} - \frac{4}{40} = \frac{9}{40}$ tank/hour. - Duration: 0.5 hours. - Water filled: $$0.5 \times \frac{9}{40} = \frac{9}{80}$$ tank. 6. **Total water filled after 2.5 hours:** $$\frac{47}{60} + \frac{9}{80} = \frac{47 \times 4}{240} + \frac{9 \times 3}{240} = \frac{188}{240} + \frac{27}{240} = \frac{215}{240} = \frac{43}{48}$$ tank. 7. **Interval 3: From 2.5 hours onward (only Pipe A on)** - Filling rate: $\frac{1}{5} = \frac{24}{120}$ tank/hour. - Leak rate: $\frac{1}{10} = \frac{12}{120}$ tank/hour. - Net filling rate: $\frac{24}{120} - \frac{12}{120} = \frac{12}{120} = \frac{1}{10}$ tank/hour. 8. **Remaining tank to fill:** $$1 - \frac{43}{48} = \frac{48}{48} - \frac{43}{48} = \frac{5}{48}$$ tank. 9. **Time to fill remaining tank:** $$t = \frac{\text{remaining volume}}{\text{net rate}} = \frac{\frac{5}{48}}{\frac{1}{10}} = \frac{5}{48} \times 10 = \frac{50}{48} = \frac{25}{24} \approx 1.0417 \text{ hours}$$ 10. **Total time to fill the tank:** $$2.5 + 1.0417 = 3.5417 \text{ hours} \approx 3 \text{ hours } 32.5 \text{ minutes}$$ **Final answer:** The tank will be completely filled in approximately **3 hours and 32.5 minutes**.